This is taking from Tu's Introduction to Manifolds book. We have defined $\mathbb{R}$ as the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb{R},\phi=\operatorname{Id}_\mathbb{R}:\mathbb{R}\to\mathbb{R})$ and $\mathbb{R}'$ as the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb{R},\psi:\mathbb{R}\to\mathbb{R})$ where $\psi(x)=x^{1/3}$.
We are then instructed to show that there is a diffeomorphism between $\mathbb{R}$ and $\mathbb{R}'$, followed by a hint that it's not the identity map because it is not smooth.
My question(s):
1) Why is the identity map not a diffeomorphism/not smooth?
2) Does anyone have a hint/suggestion for an approach to finding the diffeomorphism that they are looking for?
Many thanks in advance!
Best Answer
1) When composed with the given charts, the identity map gives $(\psi \circ \operatorname{id}_\Bbb R \circ\ \phi^{-1})(x) = x^{1/3}$. This map is not smooth at $0$. Hence the identity map is not a diffeomorphism.
2) Think of a bijection that "cancels out" with the cubic root in $\psi$.